A) two
B) three
C) four
D) one
Correct Answer: B
Solution :
Ionization energy corresponding to ionization potential = -13.6 eV Photon energy incident = 12.1 eV So, the energy of electron in excited state \[=-13.6\,+12.1=-1.5\,eV\] i.e., \[{{E}_{n}}=-\frac{13.6}{{{n}^{2}}}eV\] \[-1.5\,=\frac{-13.6}{{{n}^{2}}}\] \[\Rightarrow {{n}^{2}}=\frac{-13.6}{-1.5}\approx 9\] \[\therefore n=3\] i.e., energy of electron in excited state corresponds to third orbit. The possible spectral lines are when electron jumps from orbit 3rd or 2nd; 3rd to 1st and 2nd to 1st. Thus 3 spectral lines are emitted.You need to login to perform this action.
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